Respuesta :
Answer:
The numbers are either ±13.35 and ±1.35 or ±1.35 and ±13.3
Step-by-step explanation:
Let the two numbers be 'x' and 'y'.
Given:
The squares of the sum of two numbers = 144
The sum of their squares = 180
The sum of the numbers = [tex]x+y[/tex]
The square of the sum of numbers = [tex](x+y)^2[/tex]
Square of first number = [tex]x^2[/tex]
Square of second number = [tex]y^2[/tex]
The sum of their squares = [tex]x^2+y^2[/tex]
Now, as per question:
[tex](x+y)^2=144---(1)\\\\x^2+y^2=180---(2)[/tex]
Expanding equation (1) using the formula [tex](a+b)^2=a^2+b^2+2ab[/tex]
This gives,
[tex](x+y)^2=144\\\\x^2+y^2+2xy=144\\\\\textrm{Plug in the value of }x^2+y^2\textrm{ from equation (2)}\\\\180+2xy=144\\\\2xy=144-180\\\\2xy=-36\\\\xy=-\frac{36}{2}\\\\xy=-18\\\\y=-\frac{18}{x}---(3)[/tex]
Plug in the value of 'y' from equation (3) in equation (1). This gives,
[tex]x^2+(-\frac{18}{x})^2=180\\\\x^2+\frac{324}{x^2}=180\\\\x^4+324=180x^2\\\\x^4-180x^2+324=0\\\\\textrm{On solving, we get:}\\\\x^2=178.2\ or\ x^2=1.8\\\\\textrm{Square root both sides, we get:}\\\\x=\pm13.35\ or\ x=\pm1.35[/tex]
Therefore,
[tex]y=-\frac{18}{x}=-\frac{18}{\pm13.35} = \pm1.35\ or\\\\y=-\frac{18}{x}=-\frac{18}{\pm1.35}=\pm13.3[/tex]
Therefore, the numbers are either ±13.35 and ±1.35 or ±1.35 and ±13.3
